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Figure 1. isoclines for

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Example <strong>1.</strong> Draw the <strong>isoclines</strong> with their direction markers and sketch several solution<br />

curves, including the the curve satisfying the given initial condition<br />

SOLUTIONS<br />

y ′ = 2x 2 − y, y(0) = 0.<br />

The <strong>isoclines</strong> <strong>for</strong> the given equations are the parabolas 2x 2 − y = C, here C is an arbitrary<br />

constant.<br />

y<br />

4<br />

3<br />

2<br />

1<br />

K<strong>1.</strong>5 K<strong>1.</strong>0 K0.5 0 0.5 <strong>1.</strong>0 <strong>1.</strong>5<br />

x<br />

K1<br />

<strong>Figure</strong> <strong>1.</strong> Isoclines <strong>for</strong> y ′ = 2x 2 − y<br />

y<br />

4<br />

3<br />

2<br />

1<br />

K2 K1 0 1<br />

x<br />

2<br />

K1<br />

K2<br />

<strong>Figure</strong> 2. Direction field <strong>for</strong> y ′ = 2x 2 − y


y(x)<br />

4<br />

3<br />

2<br />

1<br />

K2 K1 0 1<br />

x<br />

2<br />

K1<br />

K2<br />

<strong>Figure</strong> 3. Solutions to y ′ = 2x 2 − y<br />

Section <strong>1.</strong>4 The Approximation Method of Euler<br />

Euler’s method (or the tangent line method) is a procedure <strong>for</strong> constructing approximate<br />

solutions to an initial value problem <strong>for</strong> a first-order differential equation<br />

y ′ = f(x, y),<br />

y(x0) = y0.<br />

The main idea of this method is to construct a polygonal (broken line) approximation to<br />

the solutions of the problem (1).<br />

Assume that the the problem (1) has a unique solution ϕ(x) in some interval centered at x0.<br />

Let h be a fixed positive number (called the step size) and consider the equally spaced points<br />

xn := x0 + nh, n = 0, 1, 2, . . .<br />

The construction of values yn that approximate the solution values ϕ(xn) proceeds as follows.<br />

At the point (x0, y0), the slope of the solution to (1) is given by dy/dx = f(x0, y0). Hence, the<br />

tangent line to the curve y = ϕx at the initial point (x0, y0) is<br />

y − y0 = f(x0, y0)(x − x0), or<br />

y = y0 + f(x0, y0)(x − x0).<br />

Using the tangent line to approximate ϕx, we find that <strong>for</strong> the point x1 = x0 + h<br />

ϕ(x1) ≈ y1 := y0 + f(x0, y0)(x − x0).<br />

Next, starting at the point (x1, y1), we construct the line with slope equal to f(x1, y1). If<br />

we follow the line in stepping from x1 to x2 = x1 + h, we arrive at the approximation<br />

(1)


Repeating the process, we get<br />

ϕ(x2) ≈ y2 := y1 + f(x1, y1)(x − x1).<br />

ϕ(x3) ≈ y3 := y2 + f(x2, y2)(x − x2),<br />

ϕ(x4) ≈ y4 := y3 + f(x3, y3)(x − x3), etc.<br />

This simple procedure is Euler’s method and can be summarized by the recursive <strong>for</strong>mulas<br />

xn+1 := x0 + (n + 1)h, (2)<br />

yn+1 := yn + f(xn, yn)(x − xn), n = 0, 1, 2, . . . (3)<br />

<strong>Figure</strong> <strong>1.</strong> Polygonal-line approximation given by Euler’s method

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